A Physics student wandered and had the displacements A=85.0-m 32o South of West (vector) and B=96.0-m 36o West

Question

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A Physics student wandered and had the displacements A=85.0-m 32o South of West (vector) and B=96.0-m 36o West of North. Where will the student be with respect to the point of origin? How can he get back to his starting point?
(a) Use parallelogram method
(b) Use Laws of Sines and Cosines
(c) Use component method
(d) Get the component representation of the displacements A and B using unit vectors. Determine the final displacement using unit vectors.

Answer 1

Using compass bearings.
c = 85@238° + 96@324°
γ = 180° - 324° + 238° = 94°
c^2 = 85^2 + 96^2 - 2*85*96cos94°
c = 132.5874 m
sinα/85 = sin94°/132.5874
α = 39.98985 ≈ 40°
β = 180° - 94° - 40° = 46°
His position with respect to origin is
132.5874 m @ 284° or
132.5874 m @ 14° N of W
To get back to his starting point, travel
132.5874 m @ 104° or 14° S of E

By components,
West:
85cos32 + 96sin36 =
72.084 + 56.427 = 128.511 m
North:
- 85sin32 + 96cos36 =
- 45.043 + 77.666 = 32.622 m

Using Unit vectors:
A = 85( - 0.84805i - 0.52992j)
B = 96( - 0.58779i + 0.80902j)
C = 132.59( - 0.97030i + 0.24192j)

Answer 2

I will answer this using one method, and you can thus ceck your answer using your listed methods.

First step is to get your direction vectors all in terms of the one reference point (ie change into a 360 degree compass)
>32 degree south of west is 270-32 = 238 degreees
> 36 degree west of north is 360-36 = 324 degrees

thus your displacement vectors are
A = 85[cos(238) i + sin(238) j ]
B = 96[cos(324) i + sin(324) j ]

the resultant vector is thus the addition of vectors A and B
= 32.62 i - 128.51j

to get back to the origin, you need the resultant direction in the opposite direction (ie from B to 0, nor 0 to B)

= -32.62 i +128.51 j

getting this out of complex form
magnitude of distance= sqrt(32.62^2 + 128^2) = 132.59m
direction = tan-1(128.51/-32.62) = 284.24 degrees, or "14.24 degrees north of west"